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Lina20 [59]
3 years ago
8

I don’t get this- it’s due tmrw

Chemistry
1 answer:
Simora [160]3 years ago
7 0
Insulate the cup, use a lid
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A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
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<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH. 
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<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH. 
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<span>After the  reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:

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<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
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d1i1m1o1n [39]

Answer:

Solution for A gas has a volume of 340.0 mL at 45.90 degree celsius. What is the new temperature of the gas, in kelvin, if the volume increased to 550.0 mL.

Missing: oC. ‎| Must include: oC.

Explanation:

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